A two dimensional flow has velocities in x and y directions given by u = 2 xyt and v = -y²t, where t denotes time. The equation for streamline passing through x = 1, y = 1 is

The correct option is 𝑥𝑦² = 1.

To find the equation of the streamline, we start with the fundamental differential equation for a streamline in a two-dimensional flow:
$$\frac{dx}{u}=\frac{dy}{v}$$
where $u$ and $v$ are the velocity components in the $x$ and $y$ directions, respectively.

Given the velocity components:
$$u=2xyt$$
and
$$v=-y^{2}t$$
Substituting these expressions into the streamline equation, we get:
$$\frac{dx}{2xyt}=\frac{dy}{-y^{2}t}$$

Assuming $t\ne 0$ and $y\ne 0$, we can simplify the equation by canceling the common terms $t$ and $y$ from both denominators:
$$\frac{dx}{2x}=-\frac{dy}{y}$$
Rearranging the variables to integrate both sides:
$$\frac{dx}{x}=-2\frac{dy}{y}$$

Integrating both sides of the equation:
$$\int \frac{dx}{x}=-2\int \frac{dy}{y}$$
$$\ln (x)=-2\ln (y)+\ln (C)$$
where $\ln \left(C\right)$ is the constant of integration.

Using logarithmic properties to combine terms:
$$\ln (x)+2\ln (y)=\ln (C)$$
$$\ln (x)+\ln (y^2)=\ln (C)$$
$$\ln (x{y}^2)=\ln (C)$$
Taking the exponential of both sides yields the general streamline equation:
$$x{y}^2=C$$

To find the value of $C$ for the specific streamline passing through the point $(x,y)=(1,1)$, we substitute these values into our equation:
$$(1)\cdot (1{)}^2=C$$
$$C=1$$

Thus, the equation for the streamline passing through the point is:
$$x{y}^2=1$$